+188 In circle O, PA and PB are tangents. (The figure is not drawn to scale)
A) Prove that Triangle APO is congruent to Triangle BPO
B) Find m∠BOD for m∠AOP = 64. Explain your reasoning.
+118608 <OAP=<OBP=90 degrees A radius and the tangent it touches are at right angles to one another.
OA=OB Equal radii
OP common side
therefore
\(\triangle AOP \equiv \triangle BOP \qquad \mbox{Right angle Hypotenuse and one other side test.}\)
b)
If <AOP=64, <BOP=64 corresponding angles in congruent triangles
<BOP+<BOD = 180 adjacent supplementary angles
64 +<BOD = 180
<BOD = 116 degrees
+118608 <OAP=<OBP=90 degrees A radius and the tangent it touches are at right angles to one another.
OA=OB Equal radii
OP common side
therefore
\(\triangle AOP \equiv \triangle BOP \qquad \mbox{Right angle Hypotenuse and one other side test.}\)
b)
If <AOP=64, <BOP=64 corresponding angles in congruent triangles
<BOP+<BOD = 180 adjacent supplementary angles
64 +<BOD = 180
<BOD = 116 degrees
